While reading my lecture notes about Gauss' Law, here's what I found:

Gauss' Law is obeyed by a wider range of fields, than those represented by the electrostatic field. In particular, a field that is inverse square in r and not spherically symmetrical can satisfy Gauss' Law. In other words, Gauss' Law alone does not imply the symmetry of the field of a point source which is implicit in Coulomb's Law.

What exactly are they trying to say here? Could someone please explain?

  • $\begingroup$ Gauss' law $\oint \vec E\cdot d\vec S=q/\epsilon$ is always valid, even when the charges are not static? Of course, knowing $\oint \vec E\cdot d\vec S$ is not enough to deduce $\vec E$: for instance, a square box with $+q$ and $-q$ will have no net flux but the field is not zero on the sides of the box. $\endgroup$ May 9, 2018 at 17:37
  • $\begingroup$ I believe the bold quotation means that a point charge source, following Coulomb Law, implies a spherically symmetric field. However, this is just a private case satisfying Gauss Law. The more general case of charges satisfying Gauss Law do not imply symmetry. $\endgroup$
    – npojo
    May 9, 2018 at 19:16
  • $\begingroup$ @npojo you could have a moving point charge: Gauss’ law remains valid but the field will be distorted due to relativistic contraction as per goo.gl/images/4SE5fc coulomb’s law is, strictly speaking, only valid in the static case. $\endgroup$ May 9, 2018 at 21:04

1 Answer 1


Gauss's Law involves doing a surface integral of the electric field over some surface. There are generally two ways in which you can use it:

  • The usual direction: given some distribution of charge, find the electric field; or

  • The "backwards" direction: given the electric field over some surface, find the enclosed charge.

The "backwards" direction is straightforward to apply: just integrate your known field and be done. But this is almost never used, because you're almost never given the electric field outright without knowing anything about the charge. Much more commonly, you're given a charge distribution and asked to find the field.

The main issue with this is that Gauss's Law only gives you the surface integral of the field. There isn't really a well-defined general inverse operation for surface integration (and if there was a general inverse, it wouldn't be unique by any stretch, because many, many vector fields have the same surface integral over some surface). As such, the best you can usually hope for is that the surface integral reduces to a case where you can deduce what it should be without having to resort to sorting through an infinite number of equivalent vector fields to find which one is a valid representation of your particular charge distribution.

This is where symmetry comes in. If you know something about the symmetry of your problem, you can constrain the formerly-infinite set of vector fields with identical surface integrals to just one, i.e. the only field that would preserve the symmetry of your setup. If you also happened to choose your surface so that the field was constrained by symmetry to always be perpendicular or parallel to it, then your work is basically done - the surface integral will be trivially analytically doable, which allows you to get the right electric field.

In short, due to the fact that surface integration is not one-to-one, Gauss's Law will usually only tell you something useful if there's some symmetry of the system that you can exploit.

  • $\begingroup$ Could you elaborate little more on that? In particular, I don't understand how the constraints your introduced help us reduce the number of possible fields from infinitely many to one. I recently just read that the field should be exactly perpendicular to the chosen surface, but I don't understand why that has to be the case too. $\endgroup$
    – EM_1
    Feb 21 at 18:49

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